What mass of a 31.0% by mass glucose, C_6H_12O_6, solution contains 50.0 g of gl

Question

What mass of a 31.0% by mass glucose, C_6H_12O_6, solution contains 50.0 g of glucose? 15.5g 50.0 g 55.8 g 86.1 g 161 g What is the percent Na_2CO_3 by mass in a 1.56 molar aqueous solution? 14.2% 15.6% 16.5% 99.4% 0.134% If 11.0 g of naphthalene, C_10H_8, is dissolved in 112.5 g of chloroform, CHCl_3, what is the molarity of the solution? 0.0859 m 0.0912 m 0.764 m 11.7 m 0.0891 m What is the molality of ethanol, C_2H_5OH, in an aqueous solution that is 416% ethanol by mass? 0.742 0.0161 m 0.719 m 16.1 m 74.9 m What is the molality of a 10.0% by mass hydrochloric add solution? The density of the solution is 1.0474 g/ml. 3.05m 2.87 m 0.0105 m 0.100 m 2.64 m

Explanation / Answer

Solution:- (5) From the given information, mass of glucose is 50.0 g, mass percent is 31.0% . mass of solution = ?

mass percent = (mass of glucose/mass of solution) x 100

Let's plug in the values...

31.0 = (50.0/X) x 100

divide both sides by 100

0.310 = 50.0/X

on cross multiply...

0.310X = 50.0

divide both sides by 0.310

X = 50.0/0.310 = 161 g

So, the answer is 161 g. choice (E) is correct.

(6) 1.56 molal aqueous solution of Na2CO3 means 1.56 moles of Na2CO3 are dissolved per kg of water. molar mass of Na2CO3 is (23x2 + 12 + 16x3 = 46 + 12 + 48 = 106 g/mol).

mass of Na2CO3 = 1.56 mol x 106 g/mol = 165.36 g

mass of water = 1 kg = 1000 g

mass of solution = 165.36 g + 1000 g = 1165.36 g

mass percent of sodium carbonate = (165.36/1165.36) x 100 = 14.2%

So, the answer is 14.2% and correct choice is (A).

(7) molality is moles of solute per kg of solvent.

mass of solute(naphthalene) = 11.0 g

molar mass of solute = 12 x 10 + 1.01 x 8 = 120 + 8.08 = 128.08 g/mol

moles of solute = 11.0 g x 1mol/128.08 g = 0.0859 mol

mass of solvent = 112.5 g = 0.1125 kg

molality = 0.0859 mol/ 0.1125 kg = 0.764 m

So, the answer is 0.764 m and the correct choice is (C).

HW1) 42.6% aqueous solution of ethanol means 42.6 g of ethanol present in 100 g solution.

mass of water in the solution = 100 g - 42.6 g = 57.4 g = 0.0574 kg

molar mass of ethanol(C2H5OH) = 12 x 2 + 1.01 x 6 + 16 x 1 = 24 + 6.06 + 16 = 46.06 g/mol

moles of ethanol = 42.6 g x 1mol/46.06 g = 0.925 mol

molality = 0.925 mol/0.0574 kg = 16.1 m

So, the answer is 16.1 m and correct choice is (D).

HW2) 10.0% by mass HCl solution means 10.0 g of HCl is present in 100 ml solution.

we know that, mass = volume x density

mass of solution = 100 ml x 1.0474 g/ml = 104.74 g

mass of water = 104.74 g - 10.0 g = 94.74 g = 0.09474 kg

molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol

moles of HCl = 10.0 g x 1mol/36.46 g = 0.274 mol

molality of solution = 0.274 mol/ 0.09474 kg = 2.89 m and this is closer to 2.87 m so the correct choice is (B).

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