Question

The amount of solute (molar mass $ 60\text{ }g\text{ }mo{{l}^{-1}} $ ) that must be added to $180\, g$ of water so that the vapour pressure of water is lowered by $10\%$, is

• A. $30\, g$
• B. $60\, g$
• C. $120\, g$
• D. $12\, g$

Answer 1

The Correct Option is B

Relative lowering of vapour pressure is given by the formula
$ \frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{{{w}_{A}}}{{{M}_{A}}}\times \frac{{{M}_{B}}}{{{w}_{B}}} $
As vapour pressure of water is lowered by 10%.
$ \therefore $ $ \frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{10}{100} $
$ \therefore $ $ \frac{10}{100}=\frac{{{w}_{A}}}{60}\times \frac{18}{180} $
or $ {{w}_{A}}=60\text{ }g $

Answer 2

The formula for the relative lowering of vapor pressure:

\(\frac{{p^@ - p_s}}{{p^@}} = \frac{{\omega_A}}{{m_A}} \times \frac{{m_B}}{{\omega_B}}\)

Where:
- \(p^@\) is the vapor pressure of the pure solvent (water),
- \(p_s\) is the vapor pressure of the solution,
- \(\omega_A\) and \(\omega_B\) are the number of moles of solute and solvent respectively,
- \(m_A\) and \(m_B\) are the molar masses of the solute and solvent respectively.

We are given:
- The relative lowering of vapor pressure, \(\frac{{p^@ - p_s}}{{p^@}} = \frac{{10}}{{100}} = \frac{{1}}{{10}}\).
- \(m_A = 60\) g/mol (molar mass of the solute).
- \(m_B = 18\) g/mol (molar mass of water).
- \(\omega_B = 180\) g (mass of water).

Let's substitute these values into the formula and solve for \(\omega_A\):

\(\frac{{1}}{{10}} = \frac{{\omega_A / m_A}}{{\omega_B / m_B}}\)

\(\frac{{1}}{{10}} = \frac{{\omega_A / 60}}{{180 / 18}}\)

\(\frac{{1}}{{10}} = \frac{{\omega_A / 60}}{{10}}\)

Now, cross multiply:
\(10 \times \frac{{\omega_A}}{{60}} = 1 \times 10\)

\(\frac{{10 \times \omega_A}}{{60}} = 10\)

Now, multiply both sides by \(\frac{{60}}{{10}}\):
\(\omega_A = 60\)

So, the correct option is (B): 60g