#### Answer

$70.0\, mmHg$

#### Work Step by Step

Moles of water $n_{w}=\frac{mass\,of\, water}{molar\,mass}$
$=\frac{100\,g}{18.0\,g/mol}=5.56\,mol$
Moles of sucrose $n_{s}=\frac{mass\,of\, sucrose}{molar\,mass\,of\, sucrose}$
$=\frac{50.0\,g}{342\,g/mol}=0.146\,mol$
Mole fraction of water
$X_{water}=\frac{n_{w}}{n_{w}+n_{s}}=\frac{5.56}{5.56+0.146}=0.974$
Applying Raoult's law:
$P_{water}=(X_{water})(P^{0}_{water})$
$=0.974\times71.88\, mmHg=70.0\, mmHg$