Question

The mass (in grams) of glucose that should be dissolved in 100 g of water in order to produce same lowering of vapour pressure as is produced by dissolving 1 g of urea (molecular mass = 60) in 50 g of water is

Answer 1

Answer: 6.00

$\frac{\Delta P}{P_0}=X_B$

$\approx \frac{n_B}{n_A}=\frac{w_B\times m_A}{m_B\times w_A}$

${\left(\frac{w_B\times m_A}{m_B\times w_A}\right)}_{glucose}={\left(\frac{w_B\times m_A}{m_B\times w_A}\right)}_{urea}$

$\frac{w_B\times 18}{180\times 100}=\frac{1\times 18}{60\times 50}$

$w_B=6$ g.