Question

3.15 grams of solute is present in 200ml of 0.25M solution the solute may be

1.h2c2o4. 2h2o
2. HCL
3. H2SO4
4. HNO3​

Answer 1

Explanation:

Use the formula M = w ×1000/GMW × V

substitute the the values of W ;V and GMW of each option if u get molarity equal to given molarity then that is the answer ( answer HNO3)

Answer 2

Answer:

HNO3

Explanation:

Given

       3.15 grams of solute is present in 200ml of 0.25M solution

        Molarity =(weight/gram molecular weight) × (1000/Volume in ml)

        0.25M = (3.15/gmw) × (1000/ 200)

       gmw  = 63

In the given options

  gram molecular weight of  HNO3  = H + N+ 3(O)

                                                            = 1  + 14 + 3×16

                                                             = 63