Q-1
Molar mass of
#H_2SO_4=(2xx1+32+4xx16)g"/"mol=98g"/"mol#
So #0.25MH_2SO_4->0.25xx98g"/"L=24.5g"/"L#
So #2L" " 0.25MH_2SO_4# contains #24.5xx2=49gH_2SO_4#
Now 80% (w/w) #H_2SO_4# has sp.gr#1.75g"/"mL# So its 100g will have volume #100/1.75mL#. This volume contains 80g #H_2SO_4#.
49g #H_2SO_4#. will be obtained from #100/1.75xx49/80mL=35mL" "H_2SO_4# of 80%(w/w) sample.
Q-2
7% (w/w) #H_2SO_4# contains 7g #H_2SO_4# in 100g solution. This solution has density 1.4g/mL. So 100g solution will have volume #100/1.4mL#
So 200 mL of this solution will contain #7/(100/1.4)xx200g=19.6g" "H_2SO_4#
Again 100mL 2.45% (w/v) #H_2SO_4# solution will contain 2.45g#H_2SO_4#
So total mass of #H_2SO_4# present in 300mL mixture is #19.6+2.45=22.05g#. Finally mixture is diluted to a volume of 500mL.
So 1L of this diluted solution contains #22.05/500xx1000g=44.1g H_2SO_4#
Hence molarity of the diluted solution #(44.1g/L)/(98g"/"mol)=0.45M#
