(a) calculate the mass of cacl2·6h2o needed to prepare 0.125 m cacl2(aq) by using 500. g of water. (b) what mass of niso4·6h2o must be dissolved in 500. g of water to produce 0.22 m niso4(aq)?

(a) 13.7 g.

(b) 28.91 g.

Explanation:

∴ m = (no. of moles of solute)/(mass of water (kg))

∴ m = (mass/molar mass of solute)/(mass of water (kg)).

(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.

∵ 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.

(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?

∵ 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.

hey your not cool]      

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3