A 20.00 mL sample of 0.1015 M nitric acid is introduced into a flask, and water

Question

A 20.00 mL sample of 0.1015 M nitric acid is introduced into a flask, and water is added until the volume of the solution reaches 250. mL. What is the concentration of nitric acid in the final solution? 8.12 times 10^-3 M 5.08 times 10^-4M 1.27 M 0.406 M 3.25 times 10^-2 M Calculate the molality of a solution containing 14.3 g of NaCl in 42.2 g of water. 2.45 times 10^-1m 5.80 m 103 m 5.80 times 10^-4 m 2.45 times 10^-4 m. What mass of K_2CO_3 is needed to prepare 200. mL of a solution having a concentration 0.30 10.4 g 2.07 g 4.15 g 1.49 g 13.8 g Calculate the percent by mass of potassium nitrate in a solution made from 45.0 g KNO_3 and 295 mL of alcohol. The density of alcohol is 0.997 g/mL. 15.2% 13.3% 1.51 7.57% None of these In how many grams of water should 25.31 g of potassium nitrate (KNO_3) be dissolved to prepare a 0.1982 m solution? 1, 263 g 250.0 g 7, 917 1,000. 792.0 g

Explanation / Answer

Ans 1. A) 8.12 x 10-3 M

molarity = no.of moles / volume of solution in litres

0.1015 = n / 0.020

n = 0.020 x 0.1015

n = 0.00203

so the solution contains 0.00203 moles of nitric acid

molarity of 250 ml solution containing 0.00203 moles will be ,

M = 0.00203 / 0.250

M = 0.00812 M

So the moilarity of the solution will be 8.12 x 10-3 M

The other way to solve this is by using a simple formula

M1 V1 = M2 V2

subsituting the values in the formula

20 x 0.1015 = M2 x 250

M2 = ( 20 x 0.1015 ) / 250

M2 = 8.12 x 10-3 M

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