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18/09/2020 12:41 pm
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0.023 × 1022 molecules are present in 10g of a substance 'x'. The molarity of a solution containing 5 g of substance 'x' in 2 L solution is_________× 10-3.
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18/09/2020 12:44 pm
Number of mole of X = 6.022 x 1022/6.022 x 1023 = 10/Molar mass of X
So molar mass of X = 100 g
Molarity = 5/100 x 2 = 0.025 M
= 0.025 M
= 25 x 10-3
So P = 25