Answer:
(a) 13.7 g.
(b) 28.91 g.
Explanation:
- molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.
∴ m = (no. of moles of solute)/(mass of water (kg))
<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>
<em />
<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>
∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).
m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).
∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.
<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>
∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).
m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).
∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.
For the conversions
I will start with pressure
1atm=101.3kPa
x =700kPa
x=700kPa/101.3kPa
x=6.91atm
Temperature
273K+30.00C
303K
Volume
1L=1000ml
x =50ml
x=0.05L
PV=nRT
6.91*0.05=n*0.08206*303
0.3455=24.86418n
0.3455/24.86418=n
0.0138=n
number of moles = 0.0138moles
Note: 0.08206 is the gas constant in this case
