How many grams of sodium nitrate are needed to make 2.50L of 1.12m solution

 238 g of sodium nitrate are needed to make 2.50L of 1.12m solution

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,

n = moles of solute

= volume of solution in L

Now put all the given values in the formula of molality, we get

moles of = 2.8

moles of

Thus 238 g of sodium nitrate are needed to make 2.50L of 1.12m solution

thats how its supposed to be

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