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
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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
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Answer:
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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)