Astudent dissolved 4.00 g of co(no3)2 in enough water to make 100. ml of stock solution. he took 4.00 ml of the stock solution and then diluted it with water to give 275. ml of a final solution. how many grams of no3- ion are there in the final solution?

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n=

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

Let the molarity of the solution be

A students then takes 4 .00 mL of solution and dilute it to 275 ml.

(molarity after dilution)

(after dilution)

Molarity of the of solution after dilution is 0.002375 M.

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions =

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097  grams of nitrate ions are there in the final solution.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml