300 gm, `30%`(w/w) NaOH solution is mixed with 500 gm `40%`(w/w) NaOH solution. What is `%` (w/v) NaOH if density of final solution is 2 gm/mL?
A. 72.5
B. 65
C. 62.5
D. None
Correct Answer - C `V_(1)M_(1) + V_(2)M_(2) = V_(3) xx M_(3)` `(2.5 xx 1 +3 xx 0.5) = M_(3) xx 5.5` `:. 2.5 +1.5 = M_(3) xx 5.5` or `M_(3) =...
2 Answers 2 viewsCorrect Answer - c `HCl+NaOHto NaCl+H_(2)O` at t=0, number of moles = `(500xx0.1)/1000 = (200xx0.2)/1000 ` =0.05 =0.04 during neutralisation of 1 mole of NaOH by 1 mole of HCl, heat...
2 Answers 1 viewsCorrect Answer - (i) `DeltaG= V (P_(2)- P_(1)) + [DeltaH_(300) - 300 DeltaS_(300)]+ nRT ln .(P_(2))/(P_(1))` (ii) P = 26.28 atm
2 Answers 1 viewsCorrect Answer - B `0.4 " mole " NaOH = 0.4 xx 40 = 16 gm` `80%` by mass NaOH `80 gm NaOH` in 100 gm solution 16 gm NaOH in...
2 Answers 1 viewsCorrect Answer - c `M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` `M_(3)=(1xx2.5+0.5xx3)/(2.5+3)` `(4.0)/(5.5)=0.73M`
2 Answers 1 viewsCorrect Answer - b If the two solutions containing the same solute are mixed, the molarity of the resulting solution is given as : `M_(3)V_(3)=M_(1)V_(1)+M_(2)V_(2)` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/V_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` `=((3.0M)xx(25mL)+(4.0M)xx(75mL))/((25+75)mL)` =3.75 M
2 Answers 1 viewsCorrect Answer - ABD `P_0=X_AP_A^@+X_BP_B^@=2/8xx300+6/8xx500`=450 mm of Hg Now, RLVP=`(P_0-P_S)/P_0=(450-420)/450=1/15` Also, RLVP =`(("in")/("in"+N))` `1/15=((32i)/70)/((32i)/70+8) :. i=1.25=(1+(2-1)alpha)`. So `alpha=0.25` (or 25 %) `:. n_(Cl^(-))` produced `=25/100xx32/70=4/35` `:. n_(PbCl_2)` precipitated =`1/2xx4/35=2/35`
2 Answers 1 viewsCorrect Answer - B If two liquid of equal masses and different densities are mixed together then density of mixture `rho=(2rho_(1)rho_(2))/(rho_(1)+rho_(2)) = (2xx1xx2)/(1+2) = (4)/(3)`
2 Answers 1 viewsThe initial average temperature of the acid and the `=(299.95+299.75)/(2)=299.85K` rise in temperature `=(302.65-299.65)=2.80K` Heat evolved during neutralisation `=(100+1000+44)xx4.184xx2.8=2858.5J` `therefore` Enthalpy of neutralisation `=-(2858.5)/(100)xx1000xx(1)/(0.50)` `=-57.17kJ`
2 Answers 1 viewsCorrect Answer - A::B Second ionisation energy of He(g) `=+54.4 V` Maximum lines `=(n(n-1))/(2)" "," Radius"=0.529 (n^(2))/(z)A^(@)" , Time period"prop(n^(3))/(z)`
2 Answers 1 views