You can solve this problem through dimensional analysis.
First, find the molar mass of NaHCO3.
Na = 22.99 g
H = 1.008 g
C = 12.01 g
O (3) = 16 (3) g
Now, add them all together, you end with with the molar mass of NaHCO3.
22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.
After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3
200. * 1 = 200
200/ 84.008 = 2.38
Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.
Your final answer is 2.38 mol NaHCO3.
Okay I did the math and I'm guessing around 18*C
Answer:
Vapour pressure of benzene over the solution is 253 torr
Explanation:
According to Raoult's law for a mixture of two liquid component A and B-
vapour pressure of a component (A) in solution = 
vapour pressure of a component (B) in solution = 
Where
are mole fraction of component A and B in solution respectively
are vapour pressure of pure A and pure B respectively
Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr
So, vapour pressure of benzene in solution = 
= 253 torr
Cp stands for specific heat.