Question

A solution prepared by dissolving 171 mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -0.930°C. What is the molar mass of this sugar? The value of Kf is 1.86°C/m.

Answer 1

Answer:

The appropriate answer is "342 g/mol".

Explanation:

Given:

Freezing point,

= 0.93°C

Kf,

= 1.86°C/m

Mass,

= 171

As we know,

⇒ $Freezing \ point = Kf\times molality$

                     $0.93=1.86\times molality$

             $Molality=\frac{0.93}{1.86}$

                            $=0.5 \ m$

Now,

⇒ $Molality=\frac{moles \ of \ sugar}{mass \ of \ water}$

             $0.5=\frac{moles \ of \ sugar}{0.01}$

    $moles \ of \ sugar=0.0005 \ mol$

hence,

The molar mass of the sugar will be:

= $\frac{mass}{moles}$

= $\frac{171\times 10^{-3}}{0.0005}$

= $342 \ g/mol$