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Question: A solution is prepared by dissolving \(0 \cdot 6{\text{ g}}\) of urea (molar mass \( = 60{\text{ g m...

A solution is prepared by dissolving 0⋅6 g0 \cdot 6{\text{ g}} of urea (molar mass =60 g mol−1 = 60{\text{ g mo}}{{\text{l}}^{ - 1}}) and 1⋅8 g1 \cdot 8{\text{ g}} glucose (molar mass =180 g mol−1 = 180{\text{ g mo}}{{\text{l}}^{ - 1}} in 100 mL100{\text{ mL}} of water at 27∘C{27^ \circ }{\text{C}}. The osmotic pressure of the solution is:
(R=0⋅08206 L atm K−1 mol−1)\left( {{\text{R}} = 0 \cdot 08206{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}} \right)
4⋅92 atm4 \cdot 92{\text{ atm}}
1⋅64 atm1 \cdot 64{\text{ atm}}
2⋅46 atm2 \cdot 46{\text{ atm}}
8⋅2 atm8 \cdot 2{\text{ atm}}

Explanation

Solution

The pressure applied to a pure solvent so that it does not pass into the given solution by osmosis is known as the osmotic pressure. When two different solutions are mixed, the osmotic pressure of the final solution is calculated using the formula,
π=n1+n2V×RT\pi = \dfrac{{{n_1} + {n_2}}}{V} \times RT
Where, π\pi is the osmotic pressure,
n1{n_1} is the number of moles solute 11,
n2{n_2} is the number of moles of solute 22,
VV is the volume of the final solution in liters,
RR is the universal gas constant,
TT is the temperature in kelvin.

Complete step by step answer:
Step 1:
Convert the units of temperature from ∘C^ \circ {\text{C}} to K{\text{K}} using the relation as follows:
T(K)=T∘C+273T\left( {\text{K}} \right) = {T^ \circ }{\text{C}} + 273
Substitute 27∘C{27^ \circ }{\text{C}} for the temperature in ∘C^ \circ {\text{C}}. Thus,
T(K)=27∘C+273T\left( {\text{K}} \right) = {27^ \circ }{\text{C}} + 273
T(K)=300 KT\left( {\text{K}} \right) = 300{\text{ K}}
Thus, the temperature is 300 K300{\text{ K}}.
Step 2:
Convert the units of volume from mL{\text{mL}} to L{\text{L}} using the relation as follows:
1 mL=1×10−3 L1{\text{ mL}} = 1 \times {10^{ - 3}}{\text{ L}}
Thus,
V=100 ̸mL×1×10−3 L1 ̸mL=0⋅1 LV = 100{\text{ }}\not{{{\text{mL}}}} \times \dfrac{{1 \times {{10}^{ - 3}}{\text{ L}}}}{{1{\text{ }}\not{{{\text{mL}}}}}} = 0 \cdot 1{\text{ L}}
Thus, the volume is 0⋅1 L0 \cdot 1{\text{ L}}
Step 3:
Calculate the number of moles of urea using the formula as follows:
Number of moles of urea(n1)=Mass of urea(g)Molar mass of urea(g mol−1){\text{Number of moles of urea}}\left( {{n_1}} \right) = \dfrac{{{\text{Mass of urea}}\left( {\text{g}} \right)}}{{{\text{Molar mass of urea}}\left( {{\text{g mo}}{{\text{l}}^{ - 1}}} \right)}}
Substitute 0⋅6 g0 \cdot 6{\text{ g}} for the mass of urea, 60 g mol−160{\text{ g mo}}{{\text{l}}^{ - 1}} for the molar mass of urea. Thus,
Number of moles of urea(n1)=0⋅6 ̸g60 ̸g mol−1=0⋅01 mol{\text{Number of moles of urea}}\left( {{n_1}} \right) = \dfrac{{0 \cdot 6{\text{ }}\not{{\text{g}}}}}{{60{\text{ }}\not{{\text{g}}}{\text{ mo}}{{\text{l}}^{ - 1}}}} = 0 \cdot 01{\text{ mol}}
Thus, the number of moles of urea are 0⋅01 mol0 \cdot 01{\text{ mol}}.
Step 4:
Calculate the number of moles of glucose using the formula as follows:
Number of moles of glucose(n2)=Mass of glucose(g)Molar mass of glucose(g mol−1){\text{Number of moles of glucose}}\left( {{n_2}} \right) = \dfrac{{{\text{Mass of glucose}}\left( {\text{g}} \right)}}{{{\text{Molar mass of glucose}}\left( {{\text{g mo}}{{\text{l}}^{ - 1}}} \right)}}
Substitute 1⋅8 g1 \cdot 8{\text{ g}} for the mass of glucose, 180 g mol−1180{\text{ g mo}}{{\text{l}}^{ - 1}} for the molar mass of glucose. Thus,
Number of moles of glucose(n2)=1⋅8 ̸g180 ̸g mol−1=0⋅01 mol{\text{Number of moles of glucose}}\left( {{n_2}} \right) = \dfrac{{{\text{1}} \cdot {\text{8 }}\not{{\text{g}}}}}{{180{\text{ }}\not{{\text{g}}}{\text{ mo}}{{\text{l}}^{ - 1}}}} = 0 \cdot 01{\text{ mol}}
Thus, the number of moles of glucose are 0⋅01 mol0 \cdot 01{\text{ mol}}.
Step 5:
Calculate the osmotic pressure of the final solution using the formula as follows:
π=n1+n2V×RT\pi = \dfrac{{{n_1} + {n_2}}}{V} \times RT
Substitute 0.01 mol0.01{\text{ mol}} for the number of moles of urea, 0⋅01 mol0 \cdot 01{\text{ mol}} for the number of moles of glucose, 0⋅1 L0 \cdot 1{\text{ L}} for the volume of the final solution, 0⋅08206 L atm K−1 mol−10 \cdot 08206{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} for the universal gas constant, 300 K300{\text{ K}} for the temperature. Thus,
π=0⋅01 mol+0⋅01 mol0⋅1 L×0⋅08206 L atm K−1 mol−1×300 K\pi = \dfrac{{0 \cdot 01{\text{ mol}} + 0 \cdot 01{\text{ mol}}}}{{0 \cdot 1{\text{ L}}}} \times 0 \cdot 08206{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 300{\text{ K}}
π=0⋅02 ̸mol0⋅1 ̸L×0⋅08206 ̸L atm ̸K−1 ̸mol−1×300 ̸K\pi = \dfrac{{0 \cdot 02{\text{ }}\not{{{\text{mol}}}}}}{{0 \cdot 1{\text{ }}\not{{\text{L}}}}} \times 0 \cdot 08206{\text{ }}\not{{\text{L}}}{\text{ atm }}\not{{{{\text{K}}^{ - 1}}}}{\text{ }}\not{{{\text{mo}}{{\text{l}}^{ - 1}}}} \times 300{\text{ }}\not{{\text{K}}}
π=4⋅92 atm\pi = 4 \cdot 92{\text{ atm}}
Thus, the osmotic pressure of a solution prepared by dissolving 0⋅6 g0 \cdot 6{\text{ g}} of urea (molar mass =60 g mol−1 = 60{\text{ g mo}}{{\text{l}}^{ - 1}}) and 1⋅8 g1 \cdot 8{\text{ g}} glucose (molar mass =180 g mol−1 = 180{\text{ g mo}}{{\text{l}}^{ - 1}} in 100 mL100{\text{ mL}} of water at 27∘C{27^ \circ }{\text{C}} is 4⋅92 atm4 \cdot 92{\text{ atm}}.

So, the correct answer is “Option A”.

Note:
Do not use the temperature value in ∘C^ \circ {\text{C}}. Convert the temperature from∘C^ \circ {\text{C}} to K{\text{K}} using the relation that 0∘C=274 K{0^ \circ }{\text{C}} = 274{\text{ K}}. Calculate the number of moles of each solute using the relation that the number of moles is the ratio of mass to molar mass. As two solutes are mixed to form one solution, the molar concentration of the final solution depends on both the solutes.