9.0g of hydrated salted CuSO4 x H2O was dissolved in 500cm3 of water and it's concentration in moldm3 was experimentally found to be 0.072M . Calculate, the molar mass of CuSO4 x H2O and the value of x, the water of crystallization.[Cu=64 S=32 O=16 H=1]
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I don't understand it well so please can you solve it
M = 0.072 = 0.072 moles/dm^3 = 0.072 moles/L so in 500 cc (0.500 L) you will have 0.072/2 = 0.036 moles.
moles = grams/molar mass or molar mass = grams/mols. Substitute as follows:
molar mass CuSO4.xH2O = 9 g/0.036 = 250
CuSO4.xH2O = 250. CuSO4 = 160; therefore, 250-160 = 90 and since H2O = 18, then 90/18 = 5. Therefore, the hydrated salt is CuSO4.5H2O
moles = grams/molar mass or molar mass = grams/mols. Substitute as follows:
molar mass CuSO4.xH2O = 9 g/0.036 = 250
CuSO4.xH2O = 250. CuSO4 = 160; therefore, 250-160 = 90 and since H2O = 18, then 90/18 = 5. Therefore, the hydrated salt is CuSO4.5H2O
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