What would be the molarity of $ 3.0 NH_2SO_4 $ solution ?
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What would be the molarity and normality of solution prepared by dissolving $ 19.6 gm H_2SO_4 in dissolved water to prepare 800 ml solution ? (Molecular weight of H_2SO_4 is 98 gm/mole) $
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What amount of $ H_2SO_4 required to prepare 2 litre of 0.5 NH_2SO_4 solution ? (Molecular weight of H_2SO_4 is 98 gm/mole) $
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Explanation
To prepare 2 litres of 0.5 M Hâ‚‚SOâ‚„ solution, we need 0.5 x 2 = 1 mole of Hâ‚‚SOâ‚„. Since the molecular weight of Hâ‚‚SOâ‚„ is 98 g/mol, the amount of Hâ‚‚SOâ‚„ required is 1 x 98 = 98 gm.
What will be the concentration of solution prepared by dissolving 50 gm glucose in 200 gm water ?
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What is the normality of an aqueous solution of $ 0.5 Al_2(SO_4)_3 $ ?
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What will be the mole-fraction of water and NaOH respectively, when 260 gm NaOH dissolved in 1.8 kg water ? (M.W of watll Naoh = 18440)
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What would be the mole-fraction of solute in an aqueous solution of a substances heaving strength 4.5 m ?
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Explanation
$ X = { m \over 55.55 + m } $ X = mole fraction of solute m = molality = 4.5 m $ \therefore X = { 4.4 \over 55.55 + 4.5 } = 0.075 $
The density of 98% w/w $ H_2SO_4 $ solution is 1.8 gm/mole then, molarity of the solution is -
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Explanation
$ Density of 98 \% w/w H_2 SO_4 = 1.8 gm/ ml $ w = 98 gm $ \therefore d = { w + w_o \over v } \therefore V = { w+W_o \over d } $ $ \therefore V = { 100 \over 1.8 } ml molarity ( m) = {1000 \times w \over M\ times V } $ $ = { 1000 \times 98 \times 9.8 \over 98 \times 100 } $ = 18 M
Molarity and molality of an aqueous solution of $ H_2SO_4$ are 1.56 (M) and 1.8 (M) respectively; then, waqht will be the density of the solution ?
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Explanation
$ molality = { 100 \times molality \over ( 1000 \times density ) - (mol.mass of solute \times molarity ) }$
A solution is prepared from A, B, C and D mole-fraction of A, B and C are 0.1, 0.2 and 0.4 respectively then, mole-fraction of D is -
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Explanation
The sum of mole fractions of all components in a solution must be equal to 1. Given that the mole fractions of A, B, and C are 0.1, 0.2, and 0.4 respectively, the mole fraction of D can be calculated by subtracting the sum of these values from 1. (0.1 + 0.2 + 0.4 = 0.7), so the mole fraction of D is (1 - 0.7 = 0.3).
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