Question

A solution containing 36gm per $ \text{d}{{\text{m}}^{3}} $ of glucose is isotonic with a 3% (w/v) solution of a non- volatile and non-electrolyte solute. The molar mass of solute is
(A) $ 200\text{ g mo}{{\text{l}}^{-1}} $
(B) $ 150\text{ g mo}{{\text{l}}^{-1}} $
(C) $ 20\text{ g mo}{{\text{l}}^{-1}} $
(D) $ 15\text{ g mo}{{\text{l}}^{-1}} $

Answer 1

Hint: The given solution of glucose is isotonic with the non-volatile solute it means that their pressures are the same. Using ideal gas law, PV=nRT equate pressures for both glucose and the given solute to find molar mass of solute. 3% (w/v) solution means that 3gm of solute is present in 100ml of solution.

Formula used: PV = nRT, No. of moles(n) = given mass/Molar mass
Where
P is pressure
V is volume
n is no. of moles
R is universal gas constant
T is temperature.

Complete Step by Step Solution
For isotonic solution pressure is same i.e. for glucose- $ \text{P}{{\text{V}}_{\text{G}}}={{\text{n}}_{\text{G}}}\text{RT} $ and for solute- $ \text{P}{{\text{V}}_{\text{s}}}={{\text{n}}_{\text{s}}}\text{RT} $ . Equating the pressures we get, $ \dfrac{{{\text{n}}_{\text{G}}}}{{{\text{V}}_{\text{G}}}}=\dfrac{{{\text{n}}_{\text{S}}}}{{{\text{V}}_{\text{S}}}} $ (R is constant and temperature is also same for the solution)
Calculating molar mass of glucose: Formula for glucose is $ {{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}} $ , so molar mass is $ \begin{align}
  & =6\times 12+12\times 1+6\times 16 \\
 & =72+12+96 \\
 & =180 \\
\end{align} $
Calculating no. of moles of glucose $ {{\text{n}}_{\text{G}}}=\dfrac{36}{180}=0\cdot 2 $ and $ {{\text{V}}_{\text{G}}}=1\text{ d}{{\text{m}}^{3}}=1\text{ litre} $
For the solute 3% (w/v) means that 3gm of solute is present in 100ml of solution. So, $ \begin{align}
  & \dfrac{{{\text{n}}_{\text{s}}}}{{{\text{V}}_{\text{s}}}}=\dfrac{3}{\text{molar mass}\times 100\text{ ml}} \\
 & \dfrac{{{\text{n}}_{\text{s}}}}{{{\text{V}}_{\text{s}}}}=\dfrac{3}{\text{molar mass}\times \dfrac{100}{1000}\text{litre}} \\
 & \dfrac{{{\text{n}}_{\text{s}}}}{{{\text{V}}_{\text{s}}}}=\dfrac{3\times 1000}{\text{molar mass}\times 100} \\
\end{align} $
So, now equating $ \dfrac{{{\text{n}}_{\text{G}}}}{{{\text{V}}_{\text{G}}}}=\dfrac{{{\text{n}}_{\text{s}}}}{{{\text{V}}_{\text{s}}}} $ we get,
 $ \begin{align}
  & \dfrac{0\cdot 2\text{mol}}{1\text{ litre}}=\dfrac{3\text{g}\times 1000}{\text{molar mass}\times 100\text{ litre}} \\
 & \text{molar mass}=\dfrac{3\text{g}\times 1000}{0\cdot 2\text{mol}\times 100} \\
 & \text{molar mass}=\dfrac{300g}{2\text{mol}}=150\text{g mo}{{\text{l}}^{-1}} \\
\end{align} $
Hence the correct choice is (B).

Additional Information
For deriving ideal gas law, first three individual laws must be known. Boyle’s Law describes the inverse proportional relationship between pressure and volume at a constant temperature and a fixed amount of gas. Charles’s Law describes the directly proportional relationship between the volume and temperature (in Kelvin) of a fixed amount of gas, when the pressure is held constant. Avagadro’s Law states that volume of a gas is directly proportional to the amount of gas at a constant temperature and pressure. Combining these three we get Ideal Gas Law.

Note
Molar mass of a substance is the mass in grams of 1 mole of the substance. In order to solve this question one should take care of the units and one should also know how to calculate molar mass of a compound like we did for glucose. It is obtained by simply multiplying the no. of atoms of the element with its atomic mass. The conversion of milli-litre to litre should also be known.