Question

An aqueous solution of glucose boils at ${100.01}^{0}C$. The molar elevation constant for water is $0.5K/molKg$. The number of molecules of glucose in the solution containing $100g$ of water is :

• A. $6.023\times {10}^{23}$
• B. $6.023\times {10}^{22}$
• C. $12.046\times {10}^{20}$
• D. $12.046\times {10}^{23}$

Answer 1

The correct option is D $12.046\times {10}^{23}$

Here $\Delta T_b=373.16K-373.15K=0.01K$, $K_b=0.5K/molkg$, mass of water(solvent$\left(W_1\right))=100g=0.1kg$, Molar mass of glucose($M_2)=180.16g/mol$

$\Delta T_b=K_b\frac{n_{glucose}}{W_1}$

$n_{glucose}=\frac{0.01K\times 0.1kg}{0.5K/molkg}$

$n_{glucose}=2\times {10}^{-3}=\frac{N_a}{6.022\times {10}^{23}}$

$N_a=12.046\times {10}^{20}$