Calculate the amount of benzoic acid required for prparing 250ml of 0.15m solution in metanol
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Given,
Volume of solution = 250 ml
Molarity = 0.15 M
Molecular mass of benzoic acid = 122gm
Amount if benzoic acid required,w = ?
We know,
Molarity = n/V
molarity =( w/ molar mass ) x 1000/v
= M x molar mass x v/1000
= 0.15 x 122 x 250/1000
= 4.575 gm
Volume of solution = 250 ml
Molarity = 0.15 M
Molecular mass of benzoic acid = 122gm
Amount if benzoic acid required,w = ?
We know,
Molarity = n/V
molarity =( w/ molar mass ) x 1000/v
= M x molar mass x v/1000
= 0.15 x 122 x 250/1000
= 4.575 gm
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