Just normal

Chemistry Level 4

Suppose the wavelength of a moving electron having 6.37 × 10 − 27 6.37 × {10}^{-27} joules of kinetic energy be x x meters. Then the number of moles of Carbon atom containing x 100 − 0.95 × 10 − 8 \dfrac{x}{100} - 0.95 × {10}^{-8} molecules is found be y y . Then an aqueous solution of sulphuric acid of volume 17.4 17.4 ml containing y y moles of sulphuric acid is prepared. The normality of this solution is found to be z z .

What is the value of z z ?


The answer is 1E-30.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
May 1, 2016

K . E . = 6.37 × 10 − 27 1 2 m v 2 = 6.37 × 10 − 27 v 2 = 2 × 6.37 × 10 − 27 9.108 × 10 − 31 = 1.399 × 10 4 v = 1.183 × 10 2 λ = h m v = 6.625 × 10 − 34 9.108 × 10 − 31 × 1.183 × 10 2 = 6.14 × 10 − 6 ⟹ x 100 − 0.95 × 10 − 8 = 5.19 × 10 − 6 100 = 5.19 × 10 − 8 No. of moles of carbon = 5.19 × 10 − 8 6 × 10 23 ⟹ y = 8.7 × 10 − 33 Molarity of aq. sulphuric acid = 8.7 × 10 − 33 × 1000 17.4 = 10 − 30 2 Normality = M × n - factor = 10 − 30 2 × 2 = 10 − 30 ∴ z = 1 E − 30 \begin{aligned} K.E. & = 6.37 × {10}^{-27}\\ \dfrac{1}{2}mv^2 & = 6.37 × {10}^{-27}\\ v^2 & = \dfrac{2 × 6.37 × {10}^{-27}}{9.108 × {10}^{-31}}\\ & = 1.399 × {10}^4\\ v & = 1.183 × {10}^2\\ \\ \lambda & = \dfrac{h}{mv}\\ & = \dfrac{6.625 × {10}^{-34}}{9.108 × {10}^{-31} × 1.183 × {10}^2}\\ & = 6.14 × {10}^{-6}\\ \implies \dfrac{x}{100} - 0.95 × {10}^{-8} & = \dfrac{5.19 × {10}^{-6}}{100}\\ & = 5.19 × {10}^{-8}\\ \\ \text{No. of moles of carbon} & = \dfrac{5.19 × {10}^{-8}}{6 × {10}^{23}}\\ \implies y & = 8.7 × {10}^{-33}\\ \\ \text{Molarity of aq. sulphuric acid} & = \dfrac{8.7 × {10}^{-33} × 1000}{17.4}\\ & = \dfrac{{10}^{-30}}{2}\\ \\ \text{Normality} & = \text{M} × \text{n - factor}\\ & = \dfrac{{10}^{-30}}{2} × 2\\ & = {10}^{-30}\\ \\ \therefore z & = \boxed{1E-30} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I approximated and got 9.98E-31 But it marked me wrong :(

Suhas Sheikh - 3 years ago

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