Two elements $X$ and $Y$ have atomic weights of $14$ and $16$. They form a series of compounds $A, B, C, D$ and $E$ in which for the same amount of element $X$,the amounts of $Y$ are present in the ratio $1 : 2 : 3 : 4 : 5$. If the compound $A$ has $28$ parts by weight of $X$ and $16$ parts by weight of $Y$,then the compound $C$ will have $28$ parts by weight of $X$ and:

  • A
    $32$ parts by weight of $Y$
  • B
    $48$ parts by weight of $Y$
  • C
    $64$ parts by weight of $Y$
  • D
    $80$ parts by weight of $Y$

Explore More

Similar Questions

What weight of $BaCl_2$ is required to react with $24.4 \ g$ of sodium sulfate to produce $46.6 \ g$ of barium sulfate and $23.4 \ g$ of sodium chloride (in $g$)?

Which of the following sets of compounds illustrates the Law of Multiple Proportions?

Which of the following pairs of substances illustrates the Law of Multiple Proportions?

Match the laws given in List-$A$ with the scientists who discovered them in List-$B$.
List-$A$ List-$B$
$(1)$ Law of Multiple Proportions $(A)$ Dalton
$(2)$ Law of Conservation of Mass $(B)$ Joseph Proust
$(3)$ Law of Definite Proportions $(C)$ Lavoisier
$(D)$ Hofmann

$2 \ g$ of hydrogen combine with $16 \ g$ of oxygen to form water and with $6 \ g$ of carbon to form methane. In carbon dioxide,$12 \ g$ of carbon are combined with $32 \ g$ of oxygen. These figures illustrate the law of:

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo