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

The appropriate answer is "342 g/mol".

Step-by-step explanation:

Given:

Freezing point,

= 0.93°C

Kf,

= 1.86°C/m

Mass,

= 171

As we know,

⇒
`Freezing \ point = Kf* molality`

`0.93=1.86* molality`

`Molality=(0.93)/(1.86)`

`=0.5 \ m`

Now,

⇒
`Molality=(moles \ of \ sugar)/(mass \ of \ water)`

`0.5=(moles \ of \ sugar)/(0.01)`

`moles \ of \ sugar=0.0005 \ mol`

hence,

The molar mass of the sugar will be:

=
`(mass)/(moles)`

=
`(171* 10^(-3))/(0.0005)`

=
`342 \ g/mol`