Question

Question5 of 20If 2.891 g MgCl2 is dissolved in enough water to make 500.0 mL of solution, what is the molarity of the magnesium chloride solution? The molecular mass of magnesium chloride is 95.3 g/mol and the density of the solution is 1.22 g/mL.1.518 x 10-2 M5.782 x 10-3 M6.073 x 10-2 M0.5505 M5.782 M

Answer 1

Answer:

$M=0.0736M$

Explanation:

Hello,

In this case, the molarity is defined as the ratio between the moles of the solute and the volume of the solution in liters:

$M=\frac{mol_{solute}}{V_{solution}}$

Thus, the moles of the solute which is magnesium chloride are:

$mol_{MgCl_2}= 2.891 gMgCl_2*\frac{1molMgCl_2}{95.3gMgCl_2}=0.03034molMgCl_2$

Nonetheless, the volume of the solution should be computed by adding the mass of both calcium chloride and water as we know the density of the solution as shown below:

$m_{solution}=500.0mLH_2O*\frac{1gH_2O}{1mLH_2O} +2.891gMgCl_2=502.891g\ Solution$

Hence, the volume is:

$V_{solution}=502.891g\ Solution*\frac{1mL}{1.22g\ Solution}*\frac{1L}{1000mL}=0.4122L\ Solution$

Finally, the molarity results:

$M=\frac{0.03034mol}{0.4122L}\\\\M=0.0736M$

Best regards.

Answer 2

Answer:

the molarity of MgCl2 = 6.073 x 10^-2 M

Explanation:

Molarity = mole (n) divided by volume of solution (V) in liter = n/v

mole (n) = mass (m) divided by molecular mass (Mm) = m/Mm

n of MgCl2 = 2.891/95.3 = 0.03033 mole

v of solution = 500 ml = 0.500 L

Molarity = n/v = 0.03033/0.500 = 0.06066 M = 6.07 x 10^-2 M