Question

g . Calculate the molar concentration for each of the following solutions. (a) 1.50 g NaCl in 100.0 mL of solution (b) 1.50 g K2Cr2O7 in 100.0 mL of solution (c) 5.55 g CaCl2 in 125 mL of solution (d) 5.55 g Na2SO4 in 125 mL of solution

Answer 1

Answer:

(a) $M=0.257M$

(b) $M=0.0510M$

(c) $M =0.500M$

(d) $M= 0.391M$

Explanation:

Hello,

In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:

(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:

$M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M$

(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:

$M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M$

(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:

$M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M$

(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:

$M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M$

Best regards.