calculate molality of 1 M solution of NaNO₃ if its density is 1.25g/cm³
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Given, Moles of NaNO₃ = 1
Density = 1.25 g/cm³
Molar Mass of NaNO₃ = 23 + 14 + 16× 3 = 37+ 48 = 85 g
Mass of NaNO₃ in 1 L solution = 1 × 85 = 85 g
Density =
1.25 =
↪️Mass of 1 L solution =
Mass of water in solution = 1250 - 85 = 1165 g
Putting value of " volume" and " moles" in formula of Molarity,
Given, Moles of NaNO₃ = 1
Density = 1.25 g/cm³
Molar Mass of NaNO₃ = 23 + 14 + 16× 3 = 37+ 48 = 85 g
Mass of NaNO₃ in 1 L solution = 1 × 85 = 85 g
Density =
1.25 =
↪️Mass of 1 L solution =
Mass of water in solution = 1250 - 85 = 1165 g
Putting value of " volume" and " moles" in formula of Molarity,
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