JEE Main 2017 · 8 Apr · Shift 1 · Chemistry Q32

ChemistrySingle correct+4 / -1Question 32 of 90

Excess of NaOH (aq) was added to 100 mL of $\ce{FeCl3}$ (aq) resulting into 2.14 g of $\ce{Fe(OH)3}$ . The molarity of $\ce{FeCl3}$ (aq) is :

(Given molar mass of Fe = 56 g mol $^{-1}$ and molar mass of Cl = 35.5 g mol $^{-1}$ )

0.2 M

0.3 M

0.6 M

1.8 M

Answer and worked solution

Official answer: A

$\ce{Fe(OH)3}$ has molar mass 107 g mol⁻¹, so $\dfrac{2.14}{107}=0.02$ mol, which equals the moles of $\ce{FeCl3}$ . In 0.1 L that is 0.2 M.

JEE Main 2017 · 8 Apr · Shift 1, Chemistry question 32