(a) What mass of strontium nitrate (Sr(NO3)2) would be required to prepare 1.000

Question

(a) What mass of strontium nitrate (Sr(NO3)2) would be required to prepare 1.000 L of a 0.0190 M aqueous solution of this salt? 9402097X (b) Calculate the mass percent of Sr(NO3)2 in this solution, assuming the density of the solution is 1.000 kg/lL Did you recall the equation that shows how molarity is related to the number of moles of solute and the volume of the solution? Did you recall the equation that shows how the mass Check the number of significant figures. 0.40 percent of a component of a solution is defined in terms of the mass of the component in solution and the total mass of the solution?

Explanation / Answer

(a) We know that Molarity , M = ( mass/molar mass) Sr(NO3)2 / Volume of solution in L

0.0190 M = ( m / 211.6(g/mol)) / 1.000L

m = 4.02 g

(b) Mass of solution , W = Volume x density

= 1.000 L x 1.000(kg/L)

= 1.000 kg

= 1000 g

Mas of Sr(NO3)2 is , m = 4.02 g

So mass percentage = ( m / W ) x 100

= 0.402 %

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