find the %w/w of 2molal NaOH solution in temperature of 289K
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your answer is in the attachment.
assuming NaOH is dissolved in water,
molality (M) = (number of moles of solute)/(volume of solvent in litre)
therefore
2 = n/1 ( one is the volume of water)
(where n stands for number of moles)
hence number of moles = 2
percentage weight by weight = {(mass of 2 moles of NaOH)/(mass of 1 litre water) } * 100
= {(2 * 40)/1000} * 100
= 8%
assuming NaOH is dissolved in water,
molality (M) = (number of moles of solute)/(volume of solvent in litre)
therefore
2 = n/1 ( one is the volume of water)
(where n stands for number of moles)
hence number of moles = 2
percentage weight by weight = {(mass of 2 moles of NaOH)/(mass of 1 litre water) } * 100
= {(2 * 40)/1000} * 100
= 8%
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