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Question: If \(20\;g\) of a solute was dissolved in \(500\;mL\) of water and osmotic pressure of the solution ...

If 20  g20\;g of a solute was dissolved in 500  mL500\;mL of water and osmotic pressure of the solution was found to be 600mm  of  Hg600mm\;of\;Hg at 15C{{15}^\circ}C , then molecular weight of the solute is:
A) 1000  1000\;
B) 1200  1200\;
C) 1400  1400\;
D) 1800  1800\;

Explanation

Solution

We can find the molecular weight of the solute by using the Law of osmotic pressure which explains the osmotic pressure of a solute as the product of molar concentration, universal gas constant and the temperature of the solute. The value of the gas constant will be taken as per the unit of pressure and volume.

Complete answer:
Let us note down the given data and data we need to find;
Mass of the solute (in gg ) m=20gm=20g
Volume of the solute (in LL ) V=500mL=0.5LV=500mL=0.5L
Osmotic pressure of the solute (in mm  of  Hgmm\;of\;Hg ) p=600mm  of  Hgp=600mm\;of\;Hg
Temperature of solute (in KK ) T=15C=288KT=15{}^\circ C=288K
Value of Universal Gas constant (in LatmK1mol1LatmK^{-1}mol^{-1} ) R=0.0821LatmK1mol1R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}}
Molar concentration of solute C=?C=?
Moles of the solute (in   mol\;mol ) n=?n=?
Molecular weight of solute (in gg ) M=?M=?
Here, we are given the pressure in terms of mm  of  Hgmm\;of\;Hg . We know that the standard value
760mm  of  Hg=1atm760mm\;of\;Hg=1atm
Thus, the given pressure can be converted to atmospheric pressure as
p=600760=0.7895atmp=\dfrac{600}{760}=0.7895atm
Now, the osmotic pressure can be calculated by the formula
p=CRTp=CRT
Substituting the given values,
0.7895=C×0.0821×2880.7895=C\times 0.0821 \times 288
C=0.0334molL1\therefore C=0.0334mol{{L}^{-1}}
Now, the molar concentration can be defined as the ratio of the moles of the solute to the volume of the solute
C=nVC=\dfrac{n}{V}
Substituting the provided values;
0.0334=n0.50.0334=\dfrac{n}{0.5}
n=0.0167moln=0.0167mol
Now, the moles of the solute is the given mass of the solute per molar mass of the solute.
n=mMn=\dfrac{m}{M}
Substituting the obtained values,
0.0167=20M0.0167=\dfrac{20}{M}
M1200gM\simeq 1200g

Hence, the correct answer is (B)(B)

Note: We should always select the value of universal gas constant by considering the unit in which the pressure and volume are given, as it varies with the units. We should remember the value of gas constant in two standard units i.e. R=0.0821LatmK1mol1R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}} and R=8.31JK1mol1R=8.31J{{K}^{-1}}mo{{l}^{-1}} . We also have the value of gas constant in the unit of height of mercury, but as it is not given we convert it to atmospheric pressure, as the value of gas constant in terms of atm. pressure is to be remembered.