Question

How many grams of NaI would be used to produce a 23.0 M solution with a volume of 1.50 L?

Answer 1

First question:
[(3.5mol CaCl2) / 1.0 L] * 2.0 L = 7.0mol CaCl2
7.0mol CaCl2 * (110.98g CaCl2 / 1molCaCl2) = 780g CaCl2

Second question:
(Does this mean 5.00 x 102ml = 510mL solution?)
We need to convert the solution to liters because molarity is in units of mol/L:
510mL * (1L / 1000mL) = 0.540L
Convert the mass of KI to moles:
249g KI * (1mol KI / 166.0g KI) = 1.50mol KI
1.50mol KI / 0.540L = 2.78 M

Third question:
1.5L * (2.5mol / 1.0L) = 3.75mol LiF; keeping with sig. figs.: 3.8mol LiF

Fourth question:
2.0M solution of NaI = (2.0molNaI / 1 L) So since it already has 1L volume, we need only to convert the moles to grams:
2.0mol NaI * (149.89g NaI / 1mol NaI) = 0.013g NaI

Fifth question:
(Does this mean 2.50 x 102cm3 = 255cm3 solution?)
Note that 1cm3 = 1 mL so we convert the cm3 to Liters:
255cm3 * (255mL / 255cm3) * (1L / 1000mL) = 0.255L of solution.
Convert the mass of C6H12o6 to moles:
45.0g C6H12O6 * (1mol C6H12O6 / 180.16g C6H12O6) = 0.250mol C6H12O6.
Now: 0.250molC6H12O6 / 0.255L = 0.980M solution.

Sixth question:
2.50L * [ (3.5mol Sr(NO3)2) / 1L] = 8.75mol but with sig. figs. it is 8.8mol Sr(NO3)2.

Hope this helps!