Question

ws (29. How many grams of nitric acid are present in 250.0 mL of 6.70M acid solution? a. 16.8 g b. 106 g c. 335g d. 211 g 30. How many liters of water are required to prepare a 0.1590 M silver nitrate solution if 1.00 pound of silver nitrate is used? a. 8.38 L b. 16.8 L c. 7.61 L d. 36.9 L 31. Considering that calcium chloride is a strong electrolyte, what is the molarity of the chloride ions you would find in a solution prepared by mixing 8.99 g of calcium chloride with enough water to prepare 150.0 mL of solution? a. 0.540 M b. 0.270 M c. 66.00 M d. 1.08 M 32. When 25.0 mL of a 10,6 M HCl solution is diluted to 200.0 ml, what is the final molarity of this acidic solution? a. 1.33 M b. 2.65 M c. 3.98 M d. 3.53 M
#29 and 32

Answer 1

Nitric acid (HNO3) = 63.019/02) Thus Molo moleculad mars of Nitric acid (HNO - Given - molurity = 6.70M volume = 250ml to wt of HNO3 x 1000 M.wt of HNO 250 - 6.70 = wt of HNO3 x 1000 63.01 g/mol 250 = wt of HINO = 6.70 x 63.01 x 250 y 1000 = 105.54175 ~ 106 g (Ans) Given, Molarity of HCO= 10.6M. Molecular mens of rel= 36.5 g/mol volume = 25ml. C ut of hee = 10.6 x 36.5 x 25. 1000 = 9.6725 g

So, if diluted to 200 ml, then molarity = wt of till x 1000 Micet of rice 200 - M = 9.67255 36.5 ME 1.325 molef eit =) or 1.325 m or = 1.33M (Ang)