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
Answer:
[tex]M=0.0736M[/tex]
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:
[tex]M=\frac{mol_{solute}}{V_{solution}}[/tex]
Thus, the moles of the solute which is magnesium chloride are:
[tex]mol_{MgCl_2}= 2.891 gMgCl_2*\frac{1molMgCl_2}{95.3gMgCl_2}=0.03034molMgCl_2[/tex]
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:
[tex]m_{solution}=500.0mLH_2O*\frac{1gH_2O}{1mLH_2O} +2.891gMgCl_2=502.891g\ Solution[/tex]
Hence, the volume is:
[tex]V_{solution}=502.891g\ Solution*\frac{1mL}{1.22g\ Solution}*\frac{1L}{1000mL}=0.4122L\ Solution[/tex]
Finally, the molarity results:
[tex]M=\frac{0.03034mol}{0.4122L}\\\\M=0.0736M[/tex]
Best regards.
