319185 The concentration of a 100 ml solution containing X grams of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) (mol. wt. = 106) is Y M. The values of X and Y are respectively:

319186 The volume of water that must be added to a mixture of 250ml of 6M HCl and 650ml of 3M HCl to obtain 3M solution is

319187 If 1 M and 2.5 liter NaOH solution is mixed with another 0.5 M and 3 liter NaOH solution, then molarity of the resultant solution will be

319188 Read Assertion and Reason carefully and mark correct option.
Assertion :
3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
Reason :
Molar mass of hydrated oxalic acid is \(126 \mathrm{~g} \mathrm{~mol}^{-1}\).

319189 An aqueous solution of glucose is \(10 \%\) in strength. The volume in which \(1 \mathrm{~g}\) mole of it is dissolved will be

319185 The concentration of a 100 ml solution containing X grams of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) (mol. wt. = 106) is Y M. The values of X and Y are respectively:

319186 The volume of water that must be added to a mixture of 250ml of 6M HCl and 650ml of 3M HCl to obtain 3M solution is

319187 If 1 M and 2.5 liter NaOH solution is mixed with another 0.5 M and 3 liter NaOH solution, then molarity of the resultant solution will be

319188 Read Assertion and Reason carefully and mark correct option.
Assertion :
3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
Reason :
Molar mass of hydrated oxalic acid is \(126 \mathrm{~g} \mathrm{~mol}^{-1}\).

319189 An aqueous solution of glucose is \(10 \%\) in strength. The volume in which \(1 \mathrm{~g}\) mole of it is dissolved will be

319185 The concentration of a 100 ml solution containing X grams of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) (mol. wt. = 106) is Y M. The values of X and Y are respectively:

319186 The volume of water that must be added to a mixture of 250ml of 6M HCl and 650ml of 3M HCl to obtain 3M solution is

319187 If 1 M and 2.5 liter NaOH solution is mixed with another 0.5 M and 3 liter NaOH solution, then molarity of the resultant solution will be

319188 Read Assertion and Reason carefully and mark correct option.
Assertion :
3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
Reason :
Molar mass of hydrated oxalic acid is \(126 \mathrm{~g} \mathrm{~mol}^{-1}\).

319189 An aqueous solution of glucose is \(10 \%\) in strength. The volume in which \(1 \mathrm{~g}\) mole of it is dissolved will be

319185 The concentration of a 100 ml solution containing X grams of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) (mol. wt. = 106) is Y M. The values of X and Y are respectively:

319186 The volume of water that must be added to a mixture of 250ml of 6M HCl and 650ml of 3M HCl to obtain 3M solution is

319187 If 1 M and 2.5 liter NaOH solution is mixed with another 0.5 M and 3 liter NaOH solution, then molarity of the resultant solution will be

319188 Read Assertion and Reason carefully and mark correct option.
Assertion :
3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
Reason :
Molar mass of hydrated oxalic acid is \(126 \mathrm{~g} \mathrm{~mol}^{-1}\).

319189 An aqueous solution of glucose is \(10 \%\) in strength. The volume in which \(1 \mathrm{~g}\) mole of it is dissolved will be

319185 The concentration of a 100 ml solution containing X grams of \({\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\) (mol. wt. = 106) is Y M. The values of X and Y are respectively:

319186 The volume of water that must be added to a mixture of 250ml of 6M HCl and 650ml of 3M HCl to obtain 3M solution is

319187 If 1 M and 2.5 liter NaOH solution is mixed with another 0.5 M and 3 liter NaOH solution, then molarity of the resultant solution will be

319188 Read Assertion and Reason carefully and mark correct option.
Assertion :
3.1500 g of hydrated oxalic acid dissolved in water to make 250.0 mL solution will result in 0.1 M oxalic acid solution.
Reason :
Molar mass of hydrated oxalic acid is \(126 \mathrm{~g} \mathrm{~mol}^{-1}\).

319189 An aqueous solution of glucose is \(10 \%\) in strength. The volume in which \(1 \mathrm{~g}\) mole of it is dissolved will be