Do you know Chemistry? Help me please!!?

Question

A 230.mL sample of .275 M CaCl2 solution is left on a hot plate overnight; the following morning, the solution is 1.10 M. What volume of water evaportated from the .275 M CaCl2 solution?

AND/OR

A 1.42 g sample of a pure compound, with formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.

thank you!

Answer 1

1.) To figure this out, you need to know how many moles of CaCl2 there are.

.275 moles/L CaCl2 * .230 L = 0.06325 moles of CaCl2.
Take that, and divide by the new molarity:
0.06325 moles / 1.10 mole/L = 0.0575 L of solution

Thus, 230 mL - 57.5 mL = 172.5 mL of water evaporated.

2.) M2SO4 + CaCl2 -> CaSO4 + 2M and Cl2 (or 2MCl, it doesn't matter for this purpose).
Molar Mass of CaSO4 = 136.142 g/mol
1 mol of M2SO4 reacts to become 1 mol of CaSO4.
1.36 g of CaSO4 / 136.142 g/mol = 0.009990 moles CaSO4 and M2SO4

If there is 1.42g of M2SO4 and 0.009990 moles, then the molar mass must be 1.42 / 0.009990 = 142.148 g/mol
SO4 = 32.065 + 4 * 16 = 96.065
M2 = 142.148 - 96.065 = 46.083
M = 46.083/2 = 23.0415 g/mol

M must be Sodium (Na), which has an atomic mass of 23.

A faster method: the weight of both compounds is roughly equal, but there are 2 moles of M per mole of Ca. Thus, M must be roughly half the weight of Ca, and must be an alkali metal (i.e have a single positive charge when ionized, compared to Ca's 2+ charge) (group 1) - meaning Sodium.