What is the mole fraction of Compound A in a mixture containing 2.58 g of compou

Question

What is the mole fraction of Compound A in a mixture containing 2.58 g of compound A (MM = 164.9 g/mol) and 3.2 g of Compound B(MM = 153.3g/mol)?

What is the molality of Compound A in a solution containing 7.62 g of Compound A(MM = 105.5 g/mol) and 78.28 g of water (MM =18.0g/mol)

Determine the molarity when 24.9 g of Compound A (MM = 197.7 g/mol) are dissolved into enough water to make a solution of 494 mL. (Do not include units.)

What is the percent by mass of copper in an alloy containing 1.12 g of copper and 1.49 g of nickel?

What is the mass of a compound needed to make a 384.2 mL of a solution with a molarity of 7.79 M? The molar mass of the compound is 105.6 g/mol. (Do not include units)

Explanation / Answer

Given, 2.58 g of compound A (MM = 164.9 g/mol)

Moles of A = 2.58 / 164.9 = 0.0156 mol

Given, 3.2 g of compound B ( MM = 153.3 g/mol)

Moles of B = 3.2 / 153.3 = 0.021 mol

Total moles = 0.0156 + 0.021 = 0.0366 mol

Mole fraction of compound A = (Moles of A) / (Total moles) = 0.0156 / 0.0366 = 0.43

Mole fraction of compound A = 0.43

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Given, mass of compound A = 7.62 g

Moles of compound A = 7.62g / 105.5 g/mol = 0.0722 mol

Volume of solvent = 78.28 g = 78.28 mL = 0.07828 L

Molarity = Moles / Volume (in L) = 0.0722 mol / 0.07828 L = 0.922 M

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Given, mass of compound A = 24.9 g

Moles of compound A = 24.9 g / 197.7 g/mol = 0.126 mol

Volume of solvent = 494 mL = 0.494 L

Molarity = Moles / Volume (in L) = 0.126 mol / 0.494 L = 0.255 M

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Total mass of Ni and Cu = 1.49 + 1.12 = 2.61 g

Mass percent of Cu = (Mass of Cu / Total mass) x 100 = (1.12 / 2.61) x 100 = 42.9%

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Molarity = Moles / Volume(in L )

7.79 M = Moles / 0.3842

Moles = 2.99 mol

But, we know that Moles = Mass / Molar mass

given, molar mass = 105.6 g/mol

Mass = Moles x Molar mass = 2.99 x 105.6 = 315.74 g

Thus, mass of compound = 315.74 g

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