163858 How are O.P. of \(6 \%\) urea \(\pi_1\) and \(6 \%\) glucose \(\pi_2\) related to each other :

163859 0.2 molal aqueous solutions of each of \(\mathrm{NaCl}, \mathrm{BaCl}_2\) and \(\mathrm{AICl}_3\) have boiling points \(T_1, T_2\) and \(T_3\) respectively. Which of the of following is correct :

163860 The boiling point of 1 molal aq. solution of \(\mathrm{KCl}\) is: ( \(\mathrm{K}_{\mathrm{b}}\) for water \(.52 \mathrm{k}-\mathrm{kg} / \mathrm{mole}\) )

163861 The vapour pressure of an aqueous solution of glucose is \(750 \mathrm{~mm}\) of \(\mathrm{Hg}\) at \(373 \mathrm{~K}\). the molality is:

163862 \(\quad \mathbf{5 0} \mathbf{g}\) glucose is present in \(\mathbf{5 5 0} \mathrm{g}\) solution (density of solution \(=1.1 \mathrm{gm} / \mathrm{ml}\) ), Calculate the \(\% \mathrm{wt} . / \mathrm{vol}\) will be :

163858 How are O.P. of \(6 \%\) urea \(\pi_1\) and \(6 \%\) glucose \(\pi_2\) related to each other :

163859 0.2 molal aqueous solutions of each of \(\mathrm{NaCl}, \mathrm{BaCl}_2\) and \(\mathrm{AICl}_3\) have boiling points \(T_1, T_2\) and \(T_3\) respectively. Which of the of following is correct :

163860 The boiling point of 1 molal aq. solution of \(\mathrm{KCl}\) is: ( \(\mathrm{K}_{\mathrm{b}}\) for water \(.52 \mathrm{k}-\mathrm{kg} / \mathrm{mole}\) )

163861 The vapour pressure of an aqueous solution of glucose is \(750 \mathrm{~mm}\) of \(\mathrm{Hg}\) at \(373 \mathrm{~K}\). the molality is:

163862 \(\quad \mathbf{5 0} \mathbf{g}\) glucose is present in \(\mathbf{5 5 0} \mathrm{g}\) solution (density of solution \(=1.1 \mathrm{gm} / \mathrm{ml}\) ), Calculate the \(\% \mathrm{wt} . / \mathrm{vol}\) will be :

163858 How are O.P. of \(6 \%\) urea \(\pi_1\) and \(6 \%\) glucose \(\pi_2\) related to each other :

163859 0.2 molal aqueous solutions of each of \(\mathrm{NaCl}, \mathrm{BaCl}_2\) and \(\mathrm{AICl}_3\) have boiling points \(T_1, T_2\) and \(T_3\) respectively. Which of the of following is correct :

163860 The boiling point of 1 molal aq. solution of \(\mathrm{KCl}\) is: ( \(\mathrm{K}_{\mathrm{b}}\) for water \(.52 \mathrm{k}-\mathrm{kg} / \mathrm{mole}\) )

163861 The vapour pressure of an aqueous solution of glucose is \(750 \mathrm{~mm}\) of \(\mathrm{Hg}\) at \(373 \mathrm{~K}\). the molality is:

163862 \(\quad \mathbf{5 0} \mathbf{g}\) glucose is present in \(\mathbf{5 5 0} \mathrm{g}\) solution (density of solution \(=1.1 \mathrm{gm} / \mathrm{ml}\) ), Calculate the \(\% \mathrm{wt} . / \mathrm{vol}\) will be :

163858 How are O.P. of \(6 \%\) urea \(\pi_1\) and \(6 \%\) glucose \(\pi_2\) related to each other :

163859 0.2 molal aqueous solutions of each of \(\mathrm{NaCl}, \mathrm{BaCl}_2\) and \(\mathrm{AICl}_3\) have boiling points \(T_1, T_2\) and \(T_3\) respectively. Which of the of following is correct :

163860 The boiling point of 1 molal aq. solution of \(\mathrm{KCl}\) is: ( \(\mathrm{K}_{\mathrm{b}}\) for water \(.52 \mathrm{k}-\mathrm{kg} / \mathrm{mole}\) )

163861 The vapour pressure of an aqueous solution of glucose is \(750 \mathrm{~mm}\) of \(\mathrm{Hg}\) at \(373 \mathrm{~K}\). the molality is:

163862 \(\quad \mathbf{5 0} \mathbf{g}\) glucose is present in \(\mathbf{5 5 0} \mathrm{g}\) solution (density of solution \(=1.1 \mathrm{gm} / \mathrm{ml}\) ), Calculate the \(\% \mathrm{wt} . / \mathrm{vol}\) will be :

163858 How are O.P. of \(6 \%\) urea \(\pi_1\) and \(6 \%\) glucose \(\pi_2\) related to each other :

163859 0.2 molal aqueous solutions of each of \(\mathrm{NaCl}, \mathrm{BaCl}_2\) and \(\mathrm{AICl}_3\) have boiling points \(T_1, T_2\) and \(T_3\) respectively. Which of the of following is correct :

163860 The boiling point of 1 molal aq. solution of \(\mathrm{KCl}\) is: ( \(\mathrm{K}_{\mathrm{b}}\) for water \(.52 \mathrm{k}-\mathrm{kg} / \mathrm{mole}\) )

163861 The vapour pressure of an aqueous solution of glucose is \(750 \mathrm{~mm}\) of \(\mathrm{Hg}\) at \(373 \mathrm{~K}\). the molality is:

163862 \(\quad \mathbf{5 0} \mathbf{g}\) glucose is present in \(\mathbf{5 5 0} \mathrm{g}\) solution (density of solution \(=1.1 \mathrm{gm} / \mathrm{ml}\) ), Calculate the \(\% \mathrm{wt} . / \mathrm{vol}\) will be :