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Question: \[1.00\text{ }g\]of non-electrolyte solute dissolved in \[50\text{ }g\] of benzene lowered the point...

1.00 g1.00\text{ }gof non-electrolyte solute dissolved in 50 g50\text{ }g of benzene lowered the point of benzene by0.40 K0.40\text{ }K. The freezing point depression constant of benzene is 5.12 Kg/mol5.12\text{ }Kg/mol. Find the molar mass of the solute.
A.236 gm/mole.236\text{ }gm/mole.
B.256 gm/mole.256\text{ }gm/mole.
C.266 gm/mole.266\text{ }gm/mole.
D.274 gm/mole.274\text{ }gm/mole.

Explanation

Solution

We know that to answer this question is based on the formula which is used to calculate the depression in the freezing point which is denoted by ΔTf=100×w×kfM×W\Delta {{T}_{f}}=\dfrac{100\times w\times {{k}_{f}}}{M\times W} where kf{{k}_{f}} is the depression in freezing point constant.

Complete step by step solution:
We are familiar with the colligative properties of the compounds from the physical chemistry concept and also terms and definitions relating to it. Now, let us see the meaning of the depression in freezing point and how it is calculated. Depression in freezing point is the decrease in the freezing point of the solvent when a non-volatile solute is added to it or in simple words it can be said that the freezing point of a solution is less than that of freezing point of the pure solvent. Temperature plays an important role here. Now, let us calculate the above-given question according to which the data given is;
Mass of the solute (non-electrolyte) =w= 1 g;~=w=\text{ }1\text{ }g; Mass of the solvent that is W= 50 g;W=\text{ }50\text{ }g;Depression in the freezing point is given and is ΔTf=5.12Kkg/mol\Delta {{T}_{f}}=5.12Kkg/mol
So, freezing point constant is given as kf=0.40k{{k}_{f}}=0.40k
Now, the depression in the freezing point is given by the formula; ΔTf=100×w×kfM×W\Delta {{T}_{f}}=\dfrac{100\times w\times {{k}_{f}}}{M\times W}
Rearranging this equation we get;
M=100×w×kfΔTf×WM=\dfrac{100\times w\times {{k}_{f}}}{\Delta {{T}_{f}}\times W} Substituting the values, we get;
M=100×1×5.120.40×50=256g/molM=\dfrac{100\times 1\times 5.12}{0.40\times 50}=256g/mol
Therefore, the correct answer is 256 g/mol256\text{ }g/mol which the molar mass of the solute .

So, the correct answer is option C.

Note:
Remember that depression in the freezing point is also known as cryoscopy which measures the lowered freezing points in liquid by dissolved substances for determination of molecular weights of solutes and several properties of the solution and the instrument used for this is called cryoscope.