Calculate the percentage of sodium in sodium aluminium fluoride (Na₃AlF₆) correct to the nearest whole number.

(F= 19,Na = 23,Al = 27).

32 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

21 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

35 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

33 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is
C

33 %
$M o l e c u l a r m a s s o f s o d i u m a l u m i n i u m f l u o r i d e ({N a}_{3} {A l F}_{6}) = 3 \times 23 + 27 + 6 \times 19 = 69 + 27 + 114 = 210$
$M o l e c u l a r m a s s o f s o d i u m a l u m i n i u m f l u o r i d e = 210 g r a m s$
$A m o u n t o f s o d i u m i n 210 g r a m s o f s o d i u m a l u m i n i u m f l u o r i d e = 69 g r a m$
$P e r c e n t a g e o f s o d i u m i n s o d i u m a l u m i n i u m f l u o r i d e = \frac{69}{210} \times 100 = 32.85 % o r 33 %$

Suggest Corrections

Similar questions

Q. The percentage of sodium in sodium aluminium fluoride

[N a_{3} A l F_{6}]

correct to the nearest whole number is:

[

F = 19; N a = 23; A l = 27

]

(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na₃AIF₆) correct to the nearest whole number. (F = 19; Na = 23; Al = 27)
(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:
2CO + O₂ $\overset{}{\to}$ 2CO₂
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.