which of the following aqueous solution has maximum freezing point
1) 0.01MNaCl
2) 0.005M C2H5OH
3) 0.005M MgI2
4) 0.01M MgSO4
Dear Student,
Depression of freezing point of a solvent is a colligative property of a solution i.e., it depends on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Therefore, more the number of solute particles, more is the depression of freezing point of the solvent. Also ionic compounds when dissolved in water, dissociate into anions and cations, so the van't Hoff factor, i must also be taken into consideration. By this, increasing the effective particle concentration in comparison to the covalent compounds.
Let's see it one by one:
For 0.01M NaCl, one cation (Na+) and one anion (Cl-). Thus there are two moles of dissolved particles per mole of compound. giving 0.02M effective particle concentration.
For 0.005M C2H5OH, one cation (C2H5+) and (OH-). Thus there are two moles of dissolved particles per mole of compound. Giving 0.01M effective particle concentration.
For 0.005 MgI2, one cation (Mg2+) and two anions (I-). Thus there are three moles of dissolved particles per mole of compound. Giving 0.015M effective particle concentration.
For 0.01M MgSO4, one cation (Mg2+) and one anion (SO42-). Thus there are two moles of dissociated particles per mole of the compound. Giving 0.02M effective particle concentration.
So, we can see that lowest depression in freezing point is of 0.005M C2H5OH as it has least effective particle concentration. Since it has least depression, it has maximum freezing point.
Hope this information cleared your doubts about the topic.
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Regards,
Depression of freezing point of a solvent is a colligative property of a solution i.e., it depends on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Therefore, more the number of solute particles, more is the depression of freezing point of the solvent. Also ionic compounds when dissolved in water, dissociate into anions and cations, so the van't Hoff factor, i must also be taken into consideration. By this, increasing the effective particle concentration in comparison to the covalent compounds.
Let's see it one by one:
For 0.01M NaCl, one cation (Na+) and one anion (Cl-). Thus there are two moles of dissolved particles per mole of compound. giving 0.02M effective particle concentration.
For 0.005M C2H5OH, one cation (C2H5+) and (OH-). Thus there are two moles of dissolved particles per mole of compound. Giving 0.01M effective particle concentration.
For 0.005 MgI2, one cation (Mg2+) and two anions (I-). Thus there are three moles of dissolved particles per mole of compound. Giving 0.015M effective particle concentration.
For 0.01M MgSO4, one cation (Mg2+) and one anion (SO42-). Thus there are two moles of dissociated particles per mole of the compound. Giving 0.02M effective particle concentration.
So, we can see that lowest depression in freezing point is of 0.005M C2H5OH as it has least effective particle concentration. Since it has least depression, it has maximum freezing point.
Hope this information cleared your doubts about the topic.
If you have any other doubts, just post it on the group and our experts will help you out as soon as possible.
Regards,