A beaker contains 1.00 x 102 grams of 0.700 M NaCl. If you transfer 50.0 grams of the solution to another beaker, what is the molarity of the solution remaining in the first beaker?

The molarity of the solution in the first beaker would continue to be .

Explanation:

Assume that the volume of the original solution is , and that of the solution was transferred to the other beaker. (Because of the in the solution, might not necessarily be .) In this question, .

The molarity of in the solution of the first beaker was before the transfer. Calculate the number of moles of in that solution:

.

The solution in the first beaker is uniform, such that the molarity of that of transferred solution would be . Calculate the number of moles of transferred to the other beaker:

.

After the transfer, the number of moles of in the first beaker would be:

.

After the transfer, the volume of the solution in the first beaker would be .

Calculate the molarity of the solution in the first beaker after the transfer:

.

In other words, moving part of a solution from one container to another would not affect the molarity of the original solution.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

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