Boiling point of water at 750 mm Hg is 96.63 degree celsius. How many sucrose is to be added to 500 g of water such that it boils at 100 degree celsius ? Molal elevation constant for water is
0.52 kg `mol^-1` .

Explanation

Given, Mass of water, `w_(1)` = 500 g
Boiling point of water `- 99,63^(@)C` (at 750 mm Hg).
Molal elevation constant, `k_(b)`, = 0.52 K kg/mol
Molar mass of sucrose `(C_(12)H_(22)O_(11)), M_(2) = (11 xx 12 + 22 xx 1 +11 xx 16) = 342 g//mol`
Elevation of boiling point ` DeltaT_(b), = (100 + 273) - (99.63 + 273) -0.37 K`
We know that,
`Delta T_(b) = (K_(b) xx 1000 xx w_(2))/(M_(2) xx w_(1))`
`0.37 = (0.52 xx 1000 xx w_(2))/(342 xx 500)`
` w_(2) = (0.37 xx 342 xx 500)/(0.52 xx 1000)`
` w_(2) = 121.67 g`
Hence, 121.67 g(approx) Sucrose is added to 500 g of water so that it boils at `100^(@)C`