Answered

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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.

Sagot :

Cricetus

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,

⇒ [tex]Freezing \ point = Kf\times molality[/tex]

                     [tex]0.93=1.86\times molality[/tex]

             [tex]Molality=\frac{0.93}{1.86}[/tex]

                            [tex]=0.5 \ m[/tex]

Now,

⇒ [tex]Molality=\frac{moles \ of \ sugar}{mass \ of \ water}[/tex]

             [tex]0.5=\frac{moles \ of \ sugar}{0.01}[/tex]

    [tex]moles \ of \ sugar=0.0005 \ mol[/tex]

hence,

The molar mass of the sugar will be:

= [tex]\frac{mass}{moles}[/tex]

= [tex]\frac{171\times 10^{-3}}{0.0005}[/tex]

= [tex]342 \ g/mol[/tex]

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